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(MATH100)[2010](f)midterm~zzheng^_10415.pdf
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Math100 Midterm
30 October, 2010
Name :
ID :
Tutorial section:
Directions: Do all four problems. You may write on both sides of the paper if necessary. You must show your work and justify your answers in order to receive full credit. You have 1 hour, from 5pm to 6pm. No calculator will be allowed.
Problem 1 2 3 4 Total:
Scores

1. (30 points) Answer the following questions. Justify your answer for full credit.
(a)
(10 points) Find the equation of the plane that passes through the three points (1, 7, 0), (3, .1, 2) and (0, 3, 3).

(b)
(5 points) Does the point (2, 1, 2) lie on this plane?

(c)
(15 points) Is this plane the tangent plane of the surface x2 + y2 + z2 =9 at (2, 1, 2)?


Solution of (a) Let A be the point (1, 7, 0), B be (3, .1, 2) and C
.. ..
be (0, 3, 3). Then AB =< 2, .8, 2 > and AC =< .1, .4, 3 >. Soa normal vector to the plane is given by
. .
.. .. ..
. .
.. .. . i j k .
..
n = AB AC = . . 2 .8 2 . . =< .16, .8, .16 > .
. .
. .1 .4 3 .

..
Hence the desired equation is given by n . <x . 1,y . 7,z . 0 >= 0. That is,
2x + y +2z =9.
Solution of (b) Yes, the point (2, 1, 2) lies on the plane in part(a) because 2(2) + 1 + 2(2) = 9.
Solution of (c) We .rst think of the given surface as the level surface of function w = g(x, y, z)= x2 + y2 + z2 at w = 9. Then the gradient vector of g is as follows:
.g .g .g
.g(x, y, z)= <,, >=< 2x, 2y, 2z>.
.x .y .z
.g(2, 1, 2) = < 4, 2, 4 >.
Since .g(2, 1, 2) =< 4, 2, 4 >=2 < 2, 1, 2 > is a vector normal to the plane 2x + y +2z = 9, the plane 2x + y +2z = 9 is the tangent plane to the given surface at the point (2, 1, 2).
2. (20 points) The equation 4x3ez + x + cos(yz) = 6 de.nes z implicitly as a function of x and y.
(a) (15 points) Use the implicit di.erentiation to .nd .z and .z .
.x .y .z .z
(b) (5 points) Evaluate and at the point (1, 1, 0).
.x .y
Solution of (a) We treat z as an implicitly de.ned function of x and
y, i.e. z = f (x, y).
. .
. . x 4x 3 e z + x + cos(yz) = . . x [ 6 ],
4e z . . x [ x 3 ] + 4x 3 . . x [ e z ] + . . x [ x ] + . . x [ cos(yz) ] = 0,
12x 2 e z + 4x 3 e z . z . x + 1 + [. sin(yz)] . . x [ yz ] = 0,
12x 2 e z + 4x 3 e z . z . x + 1 . y sin(yz) . z . x = 0,
. z 12x 2 e z + 1
. x = . 4x3ez . y sin(yz) .
. .
. . y 4x 3 e z + x + cos(yz) = . . y [ 6 ],
4x 3 . . y [ e z ] + . . y [ x ] + . . y [ cos(yz) ] = 0,
4x 3 e z . z . y + [. sin(yz)] . . y [ yz ] = 0,
. .
4x 3 e z . z . y + [. sin(yz)] z . y . y + y . z . y = 0,
. z z sin(yz)
. y = 4x3ez . y sin(yz) .

12(1)(1)+1 = . 13 .z
Solution of (b) .z (1, 1, 0) = . and (1, 1, 0) = 0.
.x 4(1)(1).(1)(0) 4 .x
3. (25 points) Consider the intersection curve of the plane x . y . 2z =0 with the surface 6(x + y) = (2x . 2y)3/2 .
(a)
(10 points) Find a vector-valued function that describes the inter-section curve.

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